Puzzle for June 5, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
Scratchpad
Help Area
Hint #1
Add D to both sides of eq.5: E – D – F + D = A + D + F + D which becomes eq.5a) E – F = A + 2×D + F Add B to both sides of eq.6: A + B + D + B = E – B – F + B which becomes eq.6a) A + 2×B + D = E – F
Hint #2
In eq.5a, replace E – F with A + 2×B + D (from eq.6a): A + 2×B + D = A + 2×D + F Subtract A and D from both sides of the above equation: A + 2×B + D – A – D = A + 2×D + F – A – D which simplifies to eq.5b) 2×B = D + F
Hint #3
Add B to both sides of eq.2: B + D + B = C + F + B which becomes 2×B + D = C + F + B In the above equation, replace 2×B with D + F (from eq.5b): D + F + D = C + F + B which becomes 2×D + F = C + F + B Subtract F from each side: 2×D + F – F = C + F + B – F which becomes 2×D = C + B which may be written as eq.2a) 2×D = B + C
Hint #4
Add D to both sides of eq.3: D + E – C + D = A + C + D which becomes 2×D + E – C = A + C + D In the above equation, replace 2×D with B + C (from eq.2a): B + C + E – C = A + C + D which becomes eq.3a) B + E = A + C + D
Hint #5
Add B and F to both sides of eq.6: A + B + D + B + F = E – B – F + B + F which becomes A + 2×B + D + F = E In the above equation, substitute 2×B for D + F (from eq.5b): A + 2×B + 2×B = E which becomes eq.6a) A + 4×B = E
Hint #6
Substitute A + 4×B for E (from eq.6a) in eq.3a: B + A + 4×B = A + C + D which becomes A + 5×B = A + C + D Subtract A and D from both sides of the above equation: A + 5×B – A – D = A + C + D – A – D which becomes eq.3b) 5×B – D = C
Hint #7
Substitute 5×B – D for C (from eq.3b) in eq.2a: 2×D = B + 5×B – D which becomes 2×D = 6×B – D Add D to both sides of the above equation: 2×D + D = 6×B – D + D which makes 3×D = 6×B Divide both sides by 3: 3×D ÷ 3 = 6×B ÷ 3 which makes D = 2×B
Hint #8
Substitute D for 2×B in eq.5b: D = D + F Subtract D from each side of the equation above: D – D = D + F – D which makes 0 = F
Hint #9
Substitute 2×B for D in eq.3b: 5×B – 2×B = C which makes 3×B = C
Hint #10
Substitute 3×B for C, 0 for F, and 2×B for D in eq.4: 3×B + 0 – A = A + 2×B – 0 which becomes 3×B – A = A + 2×B In the above equation, add A to both sides, and subtract 2×B from both sides: 3×B – A + A – 2×B = A + 2×B + A – 2×B which makes B = 2×A Divide both sides by 2: B ÷ 2 = 2×A ÷ 2 which makes ½×B = A
Hint #11
Substitute ½×B for A in eq.6a: ½×B + 4×B = E which makes 4½×B = E
Solution
Substitute ½×B for A, 3×B for C, 2×B for D, 4½×B for E, and 0 for F in eq.1: ½×B + B + 3×B + 2×B + 4½×B + 0 = 22 which simplifies to 11×B = 22 Divide both sides of the above equation by 11: 11×B ÷ 11 = 22 ÷ 11 which means B = 2 making A = ½×B = ½ × 2 = 1 C = 3×B = 3 × 2 = 6 D = 2×B = 2 × 2 = 4 E = 4½×B = 4½ × 2 = 9 and ABCDEF = 126490